Ram draws a card randomly from cards numbered 1 to 23 and keeps it back (i.e.,…

Concept: When two independent, uniformly random draws are made — with replacement — from the same set of n distinct values {1, 2, …, n}, exactly one of three outcomes happens for any pair: the first draw is greater, the second is greater, or the two are equal. By symmetry between the two draws, P(first > second) = P(second > first). Ties occur for exactly n of the n² equally likely ordered pairs, so P(tie) = n/n² = 1/n, and the remaining probability splits evenly: P(second > first) = (1 − 1/n) / 2 = (n − 1) / (2n).

Application:

1. Ram replaces the card after drawing it ('keeps it back'), so Ram's draw and Sam's draw are independent, each uniform over the 23 numbers 1 to 23 — here n = 23.

2. Total equally likely ordered outcomes = 23 × 23 = 529.

3. Outcomes where the two cards match number 23 (one for each value 1 to 23), so P(tie) = 23/529 = 1/23.

4. The remaining probability, 1 − 1/23 = 22/23, splits equally between 'Sam's card greater' and 'Ram's card greater' by symmetry, so P(Sam's card > Ram's card) = (22/23) ÷ 2 = 11/23.

Cross-check: Enumerating directly, for each card r Ram could draw (r = 1 to 23), the number of Sam's cards strictly greater than r is (23 − r); summing gives 22 + 21 + … + 1 + 0 = (22 × 23)/2 = 253 favourable ordered pairs out of 529, i.e. 253/529, which reduces (dividing numerator and denominator by 23) to 11/23 — matching the result above. Also, this value plus its mirror (Ram's card greater, also 11/23) plus the tie probability 1/23 add up to exactly 23/23 = 1, confirming the three cases correctly cover the whole sample space.

Result: The probability that Sam draws a card greater than Ram's is 11/23.