The document analyzes the probability of two men tossing a coin in succession to win a prize for being the first to obtain heads. The first person (A) has a winning probability of 2/3, while the second person (B) has a probability of 1/3. The probabilities are derived from calculating infinite geometric series based on their respective toss sequences.

1. EXAMPLE
Two men A and B 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 𝟑𝐫𝐝
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟓𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫….
P(A) = p + q q p + 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
+ ⋯
P(A) =
1
2
+
1
8
+
1
32
+ ⋯
Now we add the probability of the above infinite geometric series by using
𝑎
1−𝑟
, 𝑤ℎ𝑒𝑟𝑒 𝑎 =
1
2
𝑎𝑛𝑑 𝑟 =
1
4
P(A) =
1
2
1−
1
4
=
1
2
×
4
3
=
2
3

2. For second person
𝐏(𝐁) = 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟐𝐧𝐝
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟒𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫 𝐏(𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐡𝐞𝐚𝐝 𝐢𝐧 𝟔𝐭𝐡
𝐭𝐨𝐬𝐬) 𝐨𝐫 ….
P(B) = q p + q q q p + q q q q q p + ⋯
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
+ ⋯
P(B) =
1
4
+
1
16
+
1
64
+ ⋯
Now we add the probability of the above infinite geometric series by using
𝑎
1−𝑟
, 𝑤ℎ𝑒𝑟𝑒 𝑎 =
1
4
𝑎𝑛𝑑 𝑟 =
1
4
P(B) =
1
4
1−
1
4
=
1
4
×
4
3
=
1
3
Their respective chances of winning are =
2
3
𝑎𝑛𝑑
1
3