The document analyzes the probabilities of three men winning a prize based on a series of coin tosses, with each aiming to be the first to obtain a head. The calculations derive the winning probabilities as 4/7 for the first person, 2/7 for the second person, and 1/7 for the third person, using the concept of infinite geometric series. The probability of winning for each individual decreases as their turn to toss the coin increases.

1. EXAMPLE
Three men toss in successionfor a prize to be given to the one, who first obtain
head. What are their respective chances of winning.
Solution
Supposefirst personis A
Let 𝑃 =
1
2
(𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑑 𝑜𝑐𝑐𝑢𝑟𝑒)
𝑞 =
1
2
(𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 ℎ𝑒𝑎𝑑 𝑛𝑜𝑡 𝑜𝑐𝑐𝑢𝑟𝑒)
𝐏(𝐀) = 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟏𝐬𝐭
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟒𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟕𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫 ….
P(A) = p + q q q p + q q q q q q p + ⋯
P(A) =
1
2
+
1
2
1
2
1
2
1
2
+
1
2
1
2
1
2
1
2
1
2
1
2
1
2
+ ⋯
P(A) =
1
2
+
1
16
+
1
128
+ ⋯
Now we add the probability of the above infinite geometric series by using
𝑎
1−𝑟
, 𝑤ℎ𝑒𝑟𝑒 𝑎 =
1
2
𝑎𝑛𝑑 𝑟 =
1
8
P(A) =
1
2
1−
1
8
=
1
2
×
8
7
=
4
7
For second person B
𝐏(𝐁) = 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟐𝐧𝐝
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟓𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟖𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫….
P(B) = q p + q q q q p + q q q q q q q p + ⋯

2. P(B) =
1
2
1
2
+
1
2
1
2
1
2
1
2
1
2
+
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
+ ⋯
P(B) =
1
4
+
1
32
+
1
256
+ ⋯
Now we add the probability of the above infinite geometric series by using
𝑎
1−𝑟
, 𝑤ℎ𝑒𝑟𝑒 𝑎 =
1
4
𝑎𝑛𝑑 𝑟 =
1
8
P(B) =
1
4
1−
1
8
=
1
4
×
8
7
=
2
7
For third personC
𝐏(𝐂) = 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟑𝐫𝐝
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟔𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟗𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫….
P(C) = q q p + q q q q q p + q q q q q q q q p + ⋯
P(C) =
1
2
1
2
1
2
+
1
2
1
2
1
2
1
2
1
2
1
2
+
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
+ ⋯
P(C) =
1
8
+
1
64
+
1
512
+ ⋯
Now we add the probability of the above infinite geometric series by using
𝑎
1−𝑟
, 𝑤ℎ𝑒𝑟𝑒 𝑎 =
1
8
𝑎𝑛𝑑 𝑟 =
1
8
P(C) =
1
8
1−
1
8
=
1
8
×
8
7
=
1
7
Their respective chances of winning are =
4
7
,
2
7
𝑎𝑛𝑑
1
7